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Homework Help: Coefficient of x^3

  1. May 15, 2008 #1
    Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3


    I know how to do simple coefficients using pascalles triangle but I really don't know how to do this.
    Any help would be much appreciated.
     
  2. jcsd
  3. May 15, 2008 #2
    Write it as

    [tex](2x^2 - 3x^{-1})^3[/tex]

    From Pascal's triangle, you know how to expand

    [tex](a+b)^n[/tex]

    What can you replace with a and what can you replace with b?
     
  4. May 15, 2008 #3

    rock.freak667

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    [tex](2x^2-\frac{3}{x})^3[/tex]

    [tex] (\frac{1}{x}(2x^3-3))^3[/tex]

    How about now?
     
  5. May 15, 2008 #4
    Oh, that's a nice way of doing it :)
     
  6. May 15, 2008 #5

    rock.freak667

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    Usually (well for me), a binomial expansion is usually done with a variable and a constant.

    as for [itex](a+b)^n[/itex] is valid for [itex]|\frac{b}{a}|<1[/itex] But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.
     
  7. May 15, 2008 #6

    Dick

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    Why is it only valid in some range??? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.
     
  8. May 15, 2008 #7

    rock.freak667

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    That's what I was taught.."validity of a binomial"
     
  9. May 15, 2008 #8

    Dick

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    Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.
     
  10. May 15, 2008 #9
    Besides, we're dealing with polynomials in the case of (a+b)^n
     
  11. May 15, 2008 #10

    Dick

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    Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.
     
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